# Introduction to Reaction Mechanisms

> AP Chemistry · Unit 5 Kinetics
> Source: https://www.owlsprep.com/study/ap-chemistry-u5-introduction-to-reaction-mechanisms/

This study guide covers core AP Chemistry reaction mechanism concepts: elementary reactions, molecularity, intermediates, catalysts, the rate-determining step, and how to derive and validate rate laws from proposed mechanisms.

**Prerequisites:** Rate law and reaction order definition; Determining rate laws from experimental data; Balancing net chemical reactions

## Learning objectives

- Define reaction mechanisms and elementary reactions
- Classify molecularity of elementary reactions and write their rate laws
- Distinguish between reaction intermediates and catalysts
- Derive overall rate laws from proposed reaction mechanisms
- Validate proposed mechanisms against experimental rate laws

## Core Fundamentals of Reaction Mechanisms

A reaction mechanism describes the step-by-step sequence of bond-breaking and bond-forming events that convert starting reactants to final products at the molecular level. Unlike the overall net reaction, which only shows starting and final species, mechanisms reveal the intermediate species that form and are consumed during the reaction.

> **Notation Conventions**
>
> Each elementary step is written as a separate reaction. Single-headed arrows indicate irreversible steps, while double equilibrium arrows indicate reversible steps.

## Elementary Reactions and Molecularity

**Elementary Reaction** — A single step in a reaction mechanism that describes one actual molecular collision or rearrangement, which cannot be broken down into smaller steps. For elementary reactions only, the rate law can be written directly from stoichiometry.

Molecularity is the number of reactant particles that participate in an elementary step. There are three common classifications:

- Unimolecular: One reactant particle reacts, rate = $k[A]$, first order overall
- Bimolecular: Two reactant particles collide and react, rate = $k[A][B]$ (or $k[A]^2$ for $2A$), second order overall
- Termolecular: Three reactant particles collide simultaneously (very rare), rate = $k[A][B][C]$, third order overall

**Worked example:** Write the rate law for each elementary reaction below, and state the molecularity of each: (a) $C_4H_8(g) \rightarrow 2C_2H_4(g)$ (b) $NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g)$

1. For reaction (a): Count the number of reactant molecules, there is only 1 molecule of $C_4H_8$, so molecularity is unimolecular. For elementary reactions, the order of each reactant equals its stoichiometric coefficient, so the rate law is:
2. $$text{rate} = k[C_4H_8]$$
3. For reaction (b): There are two different reactant molecules, so molecularity is bimolecular. Each reactant has a stoichiometric coefficient of 1, so exponents are both 1, giving the rate law:
4. $$text{rate} = k[NO][O_3]$$

> **tip**
>
> On the AP exam, you can **never** write a rate law for an overall reaction from stoichiometry—only for explicitly labeled elementary reactions. If the step is not labeled elementary, you must use experimental data or the mechanism's rate-determining step.

## Reaction Intermediates and Catalysts

Two types of species do not appear in the final net reaction: reaction intermediates and catalysts. You must be able to distinguish between these two for AP exam questions.

**Reaction Intermediate** — A species produced in one early elementary step and consumed in a subsequent later step. It is not present at the start of the reaction, and never appears in the final overall rate law.

**Catalyst** — A species consumed in an early elementary step and regenerated in a later elementary step. It is present at the start of the reaction, speeds up the reaction by changing the mechanism, and is not consumed overall.

To distinguish between the two, track the order of appearance: intermediate = product first, reactant second; catalyst = reactant first, product second.

**Worked example:** The decomposition of hydrogen peroxide in the presence of bromide ion follows this mechanism: Step 1: $H_2O_2(aq) + H^+(aq) + Br^-(aq) \rightarrow HOBr(aq) + H_2O(l)$ Step 2: $HOBr(aq) + H_2O_2(aq) \rightarrow H^+(aq) + Br^-(aq) + O_2(g) + H_2O(l)$ Identify the intermediate and catalyst in this mechanism.

1. List all species by their position: $Br^-$ is a reactant in step 1 (early step) and a product in step 2 (late step). $HOBr$ is a product in step 1 and a reactant in step 2.
2. By definition, any species consumed first (reactant early) and produced later (product late) is a catalyst, so $Br^-$ is the catalyst.
3. Any species produced first (product early) and consumed later (reactant late) is an intermediate, so $HOBr$ is the reaction intermediate.

> **tip**
>
> Always check for common ions like $H^+$ that are regenerated—they are often catalysts that students misclassify as intermediates because they do not track their order of appearance.

## Rate-Determining Step and Rate Law Derivation

The overall rate of a reaction mechanism is limited by the slowest step in the mechanism, called the **rate-determining step (RDS)**. The overall rate law is exactly equal to the rate law of the RDS.

If the RDS is the first step with no intermediates, you can write the rate law directly from the RDS stoichiometry. If the RDS comes after one or more fast reversible steps, the RDS will contain an intermediate from the fast step. You must substitute the intermediate concentration using the **pre-equilibrium approximation**: the fast step reaches equilibrium quickly, so the rate of the forward step equals the rate of the reverse step.

$$k_{forward}[reactants] = k_{reverse}[intermediate] \implies [intermediate] = \frac{k_{forward}}{k_{reverse}}[reactants]$$

**Worked example:** A reaction has this proposed mechanism: Step 1 (fast, reversible): $Br_2(g) \rightleftharpoons 2Br(g)$ (intermediate $Br$) Step 2 (slow): $Br(g) + H_2(g) \rightarrow HBr(g) + H(g)$ (intermediate $H$) Step 3 (fast): $H(g) + Br_2(g) \rightarrow HBr(g) + Br(g)$ Derive the rate law for this overall reaction.

1. The RDS is the slow step (step 2), so first write its elementary rate law:
2. $$text{rate} = k_2[Br][H_2]$$
3. $Br$ is an intermediate from the fast pre-equilibrium step 1, so substitute it using the pre-equilibrium approximation:
4. $$k_1[Br_2] = k_{-1}[Br]^2 \implies [Br] = \left(\frac{k_1}{k_{-1}}[Br_2]\right)^{1/2}$$
5. Substitute $[Br]$ into the RDS rate law to get the overall rate law:
6. $$text{rate} = k_2 \left(\frac{k_1}{k_{-1}}[Br_2]\right)^{1/2} [H_2] = k [Br_2]^{1/2}[H_2]$$
7. where $k = k_2\left(\frac{k_1}{k_{-1}}\right)^{1/2}$ is the overall rate constant.

**Worked example:** The reaction of nitrogen monoxide and oxygen is $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$. A proposed mechanism is: Step 1 (fast, reversible): $NO(g) + NO(g) \rightleftharpoons N_2O_2(g)$ (intermediate $N_2O_2$) Step 2 (slow): $N_2O_2(g) + O_2(g) \rightarrow 2NO_2(g)$ (a) Identify the intermediate. (b) Confirm the net reaction matches. (c) Derive the rate law.

1. (a) $N_2O_2$ is produced in step 1 and consumed in step 2, so it is the reaction intermediate.
2. (b) Add the two steps: $NO + NO + N_2O_2 + O_2 \rightarrow N_2O_2 + 2NO_2$. Cancel $N_2O_2$ to get $2NO(g) + O_2(g) \rightarrow 2NO_2(g)$, which matches the given overall equation.
3. (c) The RDS is step 2, so write its elementary rate law:
4. $$text{rate} = k_2[N_2O_2][O_2]$$
5. Substitute the intermediate using pre-equilibrium:
6. $$k_1[NO]^2 = k_{-1}[N_2O_2] \implies [N_2O_2] = \frac{k_1}{k_{-1}}[NO]^2$$
7. Substitute to get the final rate law:
8. $$text{rate} = k[NO]^2[O_2], \quad k = \frac{k_1k_2}{k_{-1}}$$

**Check your understanding**

Test your understanding: The experimental rate law for the reaction $2A + B \rightarrow C$ is $\text{rate} = k[A]^2[B]$. Which proposed mechanism is consistent with this rate law?

1. Which mechanism is consistent?

   - A) Step 1 (slow): $2A \rightarrow D$; Step 2 (fast): $D + B \rightarrow C$
   - B) Step 1 (fast): $A + B \rightleftharpoons D$; Step 2 (slow): $D + A \rightarrow C$
   - C) Step 1 (slow): $A + B \rightarrow D$; Step 2 (fast): $D + A \rightarrow C$
   - D) Step 1 (fast): $2A \rightleftharpoons D$; Step 2 (fast): $D + B \rightarrow C$

   *Answer:* B) Step 1 (fast): $A + B \rightleftharpoons D$; Step 2 (slow): $D + A \rightarrow C$

   *Why:* Correct: Deriving the rate law for B gives $\text{rate} = k[A]^2[B]$, which matches. Other options are inconsistent: A gives second order overall, C gives second order overall, D has no defined rate-determining step.

> **tip**
>
> When asked to confirm if a proposed mechanism matches an experimental rate law, always start with the RDS, substitute intermediates, then compare. AP exam graders require this order of working for full credit.

## Common pitfalls

- **Wrong:** Writing a rate law for the overall reaction directly from its stoichiometry
  - Why it fails: Students confuse the rule for elementary reactions with overall reactions, and the rate law sometimes matches by coincidence, leading to incorrect assumptions
  - Correct: Only write a rate law from stoichiometry if the step is explicitly labeled elementary; for all overall reactions, the rate law must come from experiment or the mechanism's RDS
- **Wrong:** Misclassifying catalyst as intermediate (or vice versa) because neither appears in the net reaction
  - Why it fails: Students only remember both are canceled from the net reaction, so they mix up the classification
  - Correct: Always track order of appearance: produced first then consumed = intermediate; consumed first then produced = catalyst
- **Wrong:** Leaving an intermediate concentration term in the final overall rate law
  - Why it fails: Students forget intermediates are not starting reactants, so their concentration depends on starting reactant concentrations
  - Correct: Always substitute any intermediate that appears in the RDS using the pre-equilibrium approximation to replace it with reactant terms
- **Wrong:** Assuming the RDS must be the first step in the mechanism
  - Why it fails: Most introductory examples have RDS as the first step, so students assume this pattern always holds
  - Correct: Always use the explicitly labeled slow step as the RDS, regardless of its position in the mechanism
- **Wrong:** Adding exponents from all elementary steps to get overall reaction order
  - Why it fails: Students assume all steps contribute equally to the overall reaction rate
  - Correct: Only exponents from the RDS (after substituting all intermediates) determine the overall reaction order

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Elementary reaction rate law | Rate order = stoichiometric coefficient | Only applies to explicitly labeled elementary reactions |
| Unimolecular elementary reaction | $\text{rate} = k[A]$ | 1 reactant particle, first order overall |
| Bimolecular ($A+B$) | $\text{rate} = k[A][B]$ | 2 particles, second order overall |
| Bimolecular ($2A$) | $\text{rate} = k[A]^2$ | 2 particles, second order overall |
| Reaction intermediate | Produced early, consumed later | Never in net reaction or final rate law |
| Catalyst | Consumed early, regenerated later | Never in net reaction, not consumed overall |
| Rate-determining step rule | Overall rate = rate of slowest step | Always use this for rate law derivation |
| Pre-equilibrium approximation | $[intermediate] = \frac{k_{forward}}{k_{reverse}}[reactants]$ | Substitutes intermediates from fast pre-RDS steps |

## What's next

This sub-topic is the foundation for all further work in kinetics and reaction dynamics. Immediately after mastering this material, you will study how catalysts modify reaction mechanisms to lower activation energy, a frequent topic on both AP MCQ and FRQ. Without understanding how to identify intermediates and extract rate laws from mechanisms, you will not be able to correctly explain catalytic behavior or validate proposed mechanisms, a common high-weight FRQ task on the AP exam. This topic also connects to equilibrium concepts in Unit 6 and organic reaction mechanisms in later units of the course.

- [Multistep Reaction Energy Profile](https://www.owlsprep.com/study/ap-chemistry-u5-multistep-reaction-energy-profile/)
- [AP Chemistry Catalysis](https://www.owlsprep.com/study/ap-chemistry-u5-catalysis/)
- [Thermodynamics Overview](https://www.owlsprep.com/study/ap-chemistry-u6-overview/)

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